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Chapter 5: Q 69. (page 429)

Solve the integral.
$鈭� \frac{x^{2}}{\sqrt{x^{2} + 1}} d x$

Short Answer

Expert verified

The solution is $\frac{x}{2} 路 \sqrt{x^{2} + 1} - \frac{1}{2} \sin h^{- 1} x + C$ .

Step by step solution

01

Step 1. Given information.

The given integral is $鈭� \frac{x^{2}}{\sqrt{x^{2} + 1}} d x$ .

02

Step 2. Solve the integral.

Let $I = 鈭� \frac{x^{2}}{\sqrt{x^{2} + 1}} d x$ .

Substitute $x = \sin h y$ , $d x = \cos h y 路 d y$ .

$\begin{array}{rcl} I & = & 鈭� \frac{{(\sin h y)}^{2} 路 \cos h y}{\sqrt{{(\sin h y)}^{2} + 1}} d y \\ = & 鈭� \frac{{(\sin h y)}^{2} 路 \cos h y}{\sqrt{{(\cos h y)}^{2}}} d y \\ = & 鈭� \frac{{(\sin h y)}^{2} 路 \cos h y}{\cos h y} d y \\ = & 鈭� {(\sin h)}^{2} d y \\ = & 鈭� \frac{\cos h 2 y - 1}{2} d y \\ = & \frac{1}{4} \sin h 2 y - \frac{y}{2} + C \\ = & \frac{1}{4} 路 2 \sin h y 路 \cos h y - \frac{y}{2} + C \\ = & \frac{1}{2} \sin h y 路 \cos h y - \frac{y}{2} + C \end{array}$

03

Step 3. Simplified answer.

$I = \frac{1}{2} \sin h y 路 \cos h y - \frac{y}{2} + C$

Substitute $x = \sin h y, \cos h y = \sqrt{x^{2} + 1}$ and $y = \sin h^{- 1} x$ .

So, the required solution is:

$I = \frac{x}{2} 路 \sqrt{x^{2} + 1} - \frac{1}{2} \sin h^{- 1} x + C$

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Most popular questions from this chapter

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